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The function $(x-y)/(x+y{)}^3$ on the unit square is not Lebesgue integrable. On $0<y<x/2$, its absolute value is at least $4/(27x^2)$, whose area integral diverges logarithmically. Its two signed iterated integrals nevertheless exist, ignoring null boundary sections, and equal $1/2$ and $-1/2$. This illustrates why absolute integrability is required for Fubini's theorem.

For a lattice polytope $P$, its Ehrhart polynomial counts integer points in dilates: $L_P(n)=\#(nP\cap {\mathbb{Z}}^d)$ for nonnegative integers $n$. Ehrhart's theorem makes this counting function a polynomial. For the unit square, $L_P(n)=(n+1{)}^2$: there are $n+1$ choices for each coordinate. This turns a geometric counting problem into algebra.^[ref]

For a $d$-dimensional lattice polytope $P$ and positive integer $n$, Ehrhart reciprocity stateswith relative interior when needed. Negative input means polynomial evaluation, not a negative dilation count. For the unit square, $(1-n{)}^2=(n-1{)}^2$ counts the interior grid points of its $n$th dilation.^[ref]

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

A square lamina is a frame obtained by removing a concentric square hole from a larger square, with corresponding sides parallel. If their side lengths are $a>b>0$, the uniform border thickness is $t=(a-b)/2$ and the frame has areaFor positive integer sides $a,b$ with the same integer parity, $t$ is a positive integer, and a unit-grid frame contains exactly $4t(b+t)$ unit squares. For example, outer side five and inner side three give thickness one and sixteen unit squares. Concentricity alone would allow a rotated inner square; the parallel-side condition is what makes this a uniform lattice border.

Let $A,B$ be nonempty sets of real numbers bounded above. Their Minkowski sum is $A+B=\{a+b:a\in A,b\in B\}$, allowing every pair from the Cartesian product $A\times B$. ThenThe sum of the two suprema is an upper bound for every permitted sum. For the reverse bound, use the approximation property of a supremum to choose $a>\sup A-\epsilon /2$ and $b>\sup B-\epsilon /2$. Their sum is within $\epsilon$ of the claimed value. This reasoning does not require either supremum to be a maximum.

Separate choices matter. On $X=[0,1]$, choose the functions $f(x)=x$ and $g(x)=1-x$. Each has supremum one, but their sum at a shared input is always one:The maximum of $f$ occurs at $x=1$, while that of $g$ occurs at $x=0$. A common input cannot realize both choices. With separate inputs, choosing $x=1$ and $y=0$ gives $f(x)+g(y)=2$. The allowed pairs of output values are the entire unit square for separate inputs, but only the line segment $a+b=1$ for the shared input.

For any two bounded functions on the same nonempty domain of a function, the useful general estimate isThe inequality can be strict, as the example shows. Equality does not require attained maxima, but it does require that the two large values can be approached at compatible inputs. For example, $f(x)=g(x)=x$ on $(0,1)$ gives equality with value two, although neither function reaches its supremum. This distinction is useful when an objective function combines terms sharing a decision variable: adding the best possible termwise scores gives an upper bound, not necessarily a feasible combined score.

The unit square here is the Cartesian product $[0,1]\times [0,1]$ in the Euclidean plane, a square of side length one.

Consider the five-point discretization of the uniformly elliptic operator $-\Delta +c$ with $c\ge 0$ on the unit square, with homogeneous Dirichlet boundary conditions. Let $h=1/(n+1)$ and $1\le k,l\le n$. In the discrete sine basis, the weighted Jacobi method has iteration matrix eigenvaluesEach one-dimensional second difference contributes $4{\sin }^2(\pi kh/2)/h^2$, and the matrix diagonal is $4/h^2+c$, giving the displayed ratio. Its value is strictly between zero and two for every finite grid. Thus $0<\omega \le 1$ puts every iteration eigenvalue strictly inside the unit circle and gives convergence on every grid. At $\omega =1$, the highest-frequency factors approach minus one as $h\to 0$, so convergence holds while high-frequency smoothing is poor.